# Which phase transition requires more heat, melting or freezing, if the mass of water/ice in both cases is "35.0 g"?

Jan 22, 2017

Neither.

Recall that the freezing point IS the melting point. That means at ${0}^{\circ} \text{C}$, the melting/freezing point of water, we really have:

$q = n \Delta {H}_{\text{trs}}$,

where:

• $q$ is the heat absorbed to melt the solid, or heat released due to freezing the water.
• $\Delta {H}_{\text{trs}}$ is the enthalpy for the phase transition.
• $n$ is the $\text{mol}$s of water in this process.

So, if the masses are both $\text{35.0 g}$, then the $\text{mol}$s of water are equal in either case:

${q}_{\text{frz" = n_1DeltaH_"frz}}$

${q}_{\text{mlt" = n_2DeltaH_"mlt}}$

where ${n}_{1} = {n}_{2} = n$.

But since $\Delta {H}_{\text{mlt" = -DeltaH_"frz" = "6.02 kJ/mol}}$, we really have that $| {q}_{\text{frz"| = |q_"mlt}} |$. Therefore, both involve the same amount of heat.

Which $q$ is positive, ${q}_{\text{frz}}$ or ${q}_{\text{mlt}}$?